In $$\triangle$$ABC, D is a point on BC. If $$\frac{AB}{AC} = \frac{BD}{DC}$$, $$\angle$$B = $$75^\circ$$ and $$\angle$$C = $$45^\circ$$ then $$\angle$$BAD is equal to:

Given, $$\angle$$B = $$75^\circ$$  and  $$\angle$$C = $$45^\circ$$

In $$\triangle$$ABC,

$$\angle$$A + $$\angle$$B + $$\angle$$C = $$180^\circ$$

$$\Rightarrow$$  $$\angle$$A + $$75^\circ$$ + $$45^\circ$$ = $$180^\circ$$

$$\Rightarrow$$  $$\angle$$A + $$120^\circ$$ = $$180^\circ$$

$$\Rightarrow$$  $$\angle$$A = $$60^\circ$$

Given, $$\frac{AB}{AC} = \frac{BD}{DC}$$

AD divides the side BC in the ratio of other two sides so AD is the angular bisector $$\angle$$A.

$$\Rightarrow$$  $$\angle$$BAD = $$\frac{1}{2}\angle$$A

$$\Rightarrow$$  $$\angle$$BAD = $$\frac{60^{\circ}}{2}$$

$$\Rightarrow$$  $$\angle$$BAD = $$30^\circ$$

Hence, the correct answer is Option C